A thin semicircular rod has total charge +Q uniformly distributed Q is placed as shown A test along it: negative point charge C is equidistant from - Q and charge +q is placed at point €: (Point from all points on the rod ) on the test charge due to the point Let Fp and Fx represent the force charge and the rod respectively. than , less than, equal to the Is the magnitude = of F, greater magnitude of F ? Explain how you can tell 2 equal= to the mugnitude = Is the magnitude of the net force on +q greater than, less than, Fp? Explain_ point charge Q is placed as shown A second negative Is the magnitude = of the net electric force on +9 greater the magnitude - of the net electric than, less than, Or equal to force on +9 in part " 6? Explain.
The Correct Answer and Explanation is:
In this setup, you have a thin semicircular rod with a uniformly distributed charge +Q+Q+Q, and a negative point charge CCC is placed equidistant from the two charges on the rod. You are interested in comparing the magnitudes of the forces acting on the test charge +q+q+q, particularly the force due to the rod (FrodF_{\text{rod}}Frod) and the force due to the point charge (FpF_{\text{p}}Fp).
1. Magnitude of FrodF_{\text{rod}}Frod vs FpF_{\text{p}}Fp
- The force FrodF_{\text{rod}}Frod is the net electrostatic force due to the uniform charge distribution along the semicircular rod. Since the charge is uniformly distributed, the forces exerted by each infinitesimal segment of the rod on the test charge +q+q+q will have both radial and tangential components. The vector sum of these infinitesimal forces results in a net force FrodF_{\text{rod}}Frod, which is directed along the axis of symmetry of the semicircle. The magnitude of this force depends on the total charge +Q+Q+Q and the distance of the test charge from the center of the semicircle.
- On the other hand, FpF_{\text{p}}Fp is the electrostatic force due to the point charge CCC, which is placed at a certain distance from the test charge +q+q+q. Since the point charge CCC exerts a Coulomb force, its magnitude is governed by Coulomb’s Law, given by: Fp=14πϵ0∣qC∣r2F_{\text{p}} = \frac{1}{4\pi\epsilon_0} \frac{|qC|}{r^2}Fp=4πϵ01r2∣qC∣
where rrr is the distance between the point charge CCC and the test charge +q+q+q.
Comparing the magnitudes, the force FrodF_{\text{rod}}Frod could be larger or smaller than FpF_{\text{p}}Fp depending on the distance and charge values. However, in many cases, the rod’s force might be less than the point charge’s force because the point charge exerts a direct, strong Coulomb force, while the rod’s charge is distributed over a larger area, making its net force potentially smaller.
2. Magnitude of the Net Electric Force on +q+q+q due to the Rod vs Point Charge
The net electric force on +q+q+q will be the vector sum of FrodF_{\text{rod}}Frod and FpF_{\text{p}}Fp. The force from the rod is distributed over the semicircular configuration, and its direction is along the axis of symmetry, while the point charge force has a specific direction based on its position relative to +q+q+q.
- If the point charge CCC is placed symmetrically with respect to the rod, the magnitude of the net electric force on +q+q+q will depend on the relative strengths of FrodF_{\text{rod}}Frod and FpF_{\text{p}}Fp. The direction of the net force will be influenced by the vector sum of these two forces.
- In a typical setup, the net force on +q+q+q would be greater than either individual force (FrodF_{\text{rod}}Frod or FpF_{\text{p}}Fp) because the forces from different sources act in different directions, and the resultant magnitude of the vector sum could be larger. However, this is contingent upon the exact configuration and distances involved.
Conclusion
The magnitude of the net force on +q+q+q would likely be greater than the magnitude of either FrodF_{\text{rod}}Frod or FpF_{\text{p}}Fp individually, depending on the relative magnitudes of the forces and their directions. The net force is a vector sum, and often such a sum leads to a larger magnitude than any of the individual forces, especially if they are not aligned in the same direction.
